Topic 15 HL IB chemistry. Energetics and thermochemistry

1. Topic 15- Energetics & Thermochemistry

2. Lesson outline….
• Lesson 1- Topic 5 review ☺.
• Lesson 2- The ionic crystal lattice.
• Lesson 3- Determining lattice enthalpy using a Born-Haber cycle.
• Lesson 4- Determining lattice enthalpy using enthalpy of solution(S).
• Lesson 5- What is entropy (system & surroundings)?
• Lesson 6- Gibb’s free energy is a useful accounting tool.
• Lesson 7- Calculating ΔG values for reactions.
• Lesson 8- Topic review☺.

3. Lesson 1- Topic 5 review☺
Level 4: state the definition for specific heat
capacity.
Level 7: Calculate the minimum wavelength of
radiation needed to break the double bond in
O2.
Level 5/6: calculate ΔH using q= mcΔT, Hess’s
law and bond enthalpies.

4. Topic 5 review.
• See review questions from Pearson.
• Answers on my conference- let’s work through them together☺
*One prescribed practical- Zn powder + CuSO4(aq) (for section A)

5. 15.1- Energy cycles
Calculating lattice enthalpy ΔHϴ
lat

6. Lesson 2- The ionic crystal lattice.
Level 4: Define enthalpy of atomization, ionization, electron
affinity and lattice formation.
Level 7: Explain the link between a Born-Haber cycle and
Hess’s law.
Level 5/6: Describe what each of these processes
involves using text, equations and particle diagrams.

7. Starter- what we should know so far…
1. Write an equation to show the first ionization energy(ΔHi
ϴ) of
sodium.
2. Why has sodium got a relatively low I.E. value? Is it positive or
negative?
3. What does the ‘ϴ’ superscript mean in the ΔHϴ symbolism?
4. Write an equation for first electron affinity (ΔHe
ϴ) of chlorine.
5. Is this process endothermic or exothermic?
6. The second electron affinity for oxygen is the opposite to this (see
data book section 8). Can you theorise why this is?

8. The ionic crystal lattice.
• Having formed gaseous ions of sodium and
chloride from before (in our starter questions),
we can combine these ions in a regular
arrangement to form a crystal lattice.
• Na+ (g) + Cl- (g) NaCl(s) - ΔHϴ
lattice formation
• The reverse of this process (dissociation) is what
we quote values for in the databook (see section
18)
• NaCl(s) Na+(g) + Cl-(g) + ΔHϴ
lattice dissociation
*These values are therefore all
positive/edothermic.

9. Lattice formation enthalpies
When an ionic lattice is formed, the oppositely charged ions are attracted to each
other. The stronger the attraction, the higher the lattice formation enthalpy.
Two factors increase the attraction and therefore the lattice formation enthalpy:
 high charge
 small size.
N3–O2–F–
Li+Na+K+
increasing lattice formation enthalpy
Task- Use your databooks (section 18) to
determine which ionic lattice takes the most
energy to break up? MgCl2 Vs. NaCl

10. Let’s practice!- remember, small and highly charged is the key AND you
need to use your periodic table (page 244 Pearson)
Potassium bromide

11. Answers!- Small and charged…
49. C. Look for the ions with the greatest charge first (this has a more
significant effect on the bond strength than size), this is Mg2+ and Ca2+ (the
anions Cl- and Br- have the same relative charges). Then consider their size.
Mg2+ is smaller than Ca2+ so it has a higher mass/charge density.
50. A. Lattice enthalpy increases in magnitude with increasing charge of the
ions and decreasing ionic radius.
51. Charge difference is clearly not the issue, so it must be difference in size
of the two cations and the type of bonding involved. Ag+ must be a smaller
ion than K+ and this is because it has far more protons (47 vs. 19, despite
having one more energy level)
In terms of bonding: the bond in AgBr is closer to being covalent than ionic
because of the closeness in electronegativity between these two elements.
Covalent bonds are harder to break than purely ionic bonds.

12. Types of enthalpy change ΔHϴ
F- Formation: The enthalpy change when one mole of a compound is formed from its constituent elements in
their standard states at room temperature and pressure. (Usually exothermic)
* remember, the enthalpy of formation of elemental substancesis zero!
A- Atomization: The enthalpy change when one mole of solid atom is converted to one mole of a gaseous
atom. (Always endothermic)
B- Bonding: The enthalpy change required to break one mole of a bond within a gaseous compound. (Always
endothermic)
I- Ionization: The enthalpy change when one mole of electrons is removed from one mole of an atom in its
gaseous state. (Always endothermic 1st, 2nd….)
E- Electron affinity: The enthalpy change when one mole of electrons is added to one mole of an atom in its
gaseous state. (Always exothermic for 1st, then endothermic for 2nd)

13. Check for learning…..
1. Write an equation to represent ΔHϴ
f of MgCl2(s)
2. Write an equation to represent the first ionization energy of
sodium ΔHϴ
i ( Na).
3. Write an equation to represent the lattice formation enthalpy of
calcium chloride ΔHϴ
Lat ( CaCl2).
4. Write an equation to represent the enthalpy of atomization of
iodine ΔHϴ
atom (I2).
5. Write an equation to represent the second electron affinity of
oxygen ΔHϴ
e ( O-).
6. Is this an endothermic or endothermic process?

14. We cannot measure lattice dissociation
experimentally….
‘The endothermic process in which
the gaseous ions of a crystal are
separated to an infinite distance
from each other’ (Thus breaking all
possible bonds between the ions)
It is impossible to achieve this under
standard conditions….
Q. So how did chemists come up
with these values for ΔHϴ
lat in
section 18 of the databook?

15. Lesson 3- Determining lattice enthalpy (ΔHϴ
lat ) using a
Born-Haber cycle.
Level 4: Recall that a Born-Haber cycle is the application of
Hess’s law to the break up of ionic compounds.
Level 7: Use a Born-Haber cycle to determine the
lattice enthalpy of an ionic compound.
Level 5/6: Construct Born-Haber cycles for
group 1 and group 2 oxides and chlorides.

16. Max Born & Fritz Haber

18. Interpreting a Born-Haber cycle….

19. The answers!
45 (a) Remember, the IB refer to the dissociation of one mole of the lattice
when they refer to lattice energy, K2O(s)  2K+(g) + O2-(g).
(b) Z= enthalpy of formation of K2O(s).
2(89.2) represents the atomization energy of 2 moles of K(s)  K(g).
W= ½ the bond enthalpy of the double bond in an oxygen molecule.
X= enthalpy of ionization of 2 moles of K(g)  K+(g).
Y= enthalpy of 1st and 2nd electron affinities of O(g).
(c) ΔHϴ
lat = +361 + 2 (89.2) + (1/2 498) + (2x 419) + (-141 + 753) = + 2238.4KJ.mol-1

20. Constructing a Born-Haber cycle.
Try and construct the Born-Haber cycle for MgO(s). The only pieces of information
you need are:
The ΔHϴ
atom of Mg(s) is +148 KJ.mol-1 ,
The second ionization energy of Mg is +1451KJ.mol-1
The ΔHϴ
f (MgO)= -602KJ.mol-1 .
The rest you should be able to find in your databooks….
*Note: You can also confirm how close your answer is to the lattice enthalpy value which is in section
18☺

21. Lesson 4- Determining lattice enthalpy using enthalpy
of solution(ΔHϴ
sol) and hydration (ΔHϴ
hyd ).
Level 4: State the definitions for enthalpy of solution and
enthalpy of hydration.
Level 7: Explain why the theoretical value may not match
the experimental value.
Level 5/6: Uses Hess’s law to determine the enthalpy of
solution of NH4Cl(s) theoretically and experimentally.

22. Enthalpies of solution and hydration
The standard enthalpy of solution (DHsol
ө) is
the enthalpy change when one mole of an
ionic compound is dissolved in water to
produce aqueous ions.
Na+
(g) Na+
(aq)
The standard enthalpy of hydration (DHhyd
ө) is the enthalpy change when one mole of
gaseous ions is converted to one mole of aqueous ions.
NaCl(s) Na+
(aq) + Cl–
(aq)

23. Calculating enthalpies of solution

24. Determining enthalpy of solution
experimentally.
The enthalpy of solution ΔHϴ
sol can be determined either using a
simple experiment where you add ionic compounds to water or using an
online simulator for these processes.
Where ΔHϴ
sol = ΔHϴ
lat + ΔHϴ
hyd
See handout from RSC ‘coffee can’ experiment.

26. 15.2- Entropy & spontaneity

27. Lesson 5- What is entropy?
Level 4: Calculate the entropy change of a chemical system
(ΔS) from given Sϴ values.
Level 7: Explain how an endothermic reaction can be
considered spontaneous even though it needs heat
from the surroundings?!
Level 5/6: Predict whether a change will result in an
increase or decrease in entropy by considering the
states of reactants and products.

28. Starter- What is entropy (S)? (demo permanganate crystals in warmed water)
First law of thermochemistry/thermodynamics- Law
of conservation of energy (kinetic energy of
reactants, heat energy of system and surroundings)
i.e. +/- ΔH
Second law- System’s particles tend to disorder i.e.
increase in entropy i.e. + ΔS e.g. the big bang theory.
*Note: you must consider the system and the
surroundings for these laws to be true.

29. What is a spontaneous reaction?

30. Calculating entropy changesStandard entropy changes for any chemical reaction or physical change can be
calculated using the following simple expression:
Remember the following points:
 entropies of elements are not zero like DHf values, so they should be included
in calculations.
 the units of entropy, S, are J K–1 mol–1
DS = SSө
products – SSө
reactants

31. *Note: For Q60 & 61, you need Sϴ = H2(g) = +131, N2(g) = +191, NH3(g) = +193, graphite(C) = +5.7, CH4(g)= +186J.K-1.mol-1

32. 54. B. 2 moles of gaseous molecules to 2 moles of new gaseous molecules.
55. C. I involves a decrease in entropy of the system from gas to liquid, the others involve an increase in the entropy of
the system.
56. A. solid to gas, increase in entropy. Also, this physical change requires heat from the surroundings so it’s
endothermic.
57. D. gaseous product produced from non- gaseous reactants.
58. (a) 4 (moles) gaseous molecule to 2 (moles) gaseous molecules. –ΔS.
(b) The number of gaseous molecules on either side hasn’t changed but, overall, the change has been from 7 moles
of reactants molecules to 5 moles of product molecules. –ΔS.
(c) Ligand displacement (SCN- for H2O) to give many more product molecules than reactant. +ΔS.
59.
60. DS = SSө
products – SSө
reactants
= 2 (193) – (191+ 3 (131)= -198 J.K-1.mol-1
61. DS = SSө
products – SSө
reactants
= 186- (5.7 + 2(131)= -82 J.K-1.mol-1

33. Calculating the total entropy change (system & surroundings)

34. In summary (taken from Pearson pg. 255)
An endothermic reaction can be spontaneous if the increase in entropy of the system is larger than the
decrease in entropy of the surrounding particles (due to heat loss ‘ΔH’)

35. Lesson 6- Gibb’s free energy is a useful accounting tool.
Level 4: State the equation which relates Gibb’s free
energy, enthalpy and entropy.
Level 7: Deduce the boiling point of water by calculation
and describe any assumptions made.
Level 5/6: Use the Gibb’s equation to predict the
feasibility of a reaction when temperature is changed.

36. Starter- Josiah Willard Gibbs (1839-1903)
https://www.youtube.com/watch?v=Fms2JkmVBL0
Watch to 1.52min

37. (see data book)☺
*Task- You need to be able to explain each of these using the Gibb’s equation, take your time…. (from Pearson pg.
258) (see this excellent video for review https://www.youtube.com/watch?v=8N1BxHgsoOw&t=333s

40. Section 12

42. Lesson 7- Calculating ΔG values for reactions.
Level 4: State the equations for calculating ΔH, ΔS and ΔG
for a reaction.
Level 7: Explain the relationship between Gibb’s free
energy and entropy for an equilibrium system.
Level 5/6: Carry out calculations for ΔG for reactions
which go to completion and for equilibrium reactions.

43. Starter- Remember

44. Answers!

45. Gibb’s free calculations
1. Calculating ΔGϴ without considering temperature:
2. Calculating ΔG at various temperatures:

46. 1.Ignoring temperat.:

47. 2. Various temperatures:
Remember:
And: From Pearson pg.260.

49. More practice!

50. Note: Since the enthalpyofformation ofall of the
reactants equals zero (all elemental),the enthalpyfor
this reaction = the enthalpyofformation ofethanol
which is -278 KJ.mol-1

51. Gibb’s free energy and equilibrium.
• When ΔGϴ
reaction is below -30 kJ.mol-1
a reaction is complete()
• When ΔGϴ
reaction is between -30 and 0
kJ.mol-1, there is an equilibrium
mixture with products predominating

52. Taken from Pearson pg 263 and the HL
midterm paper 2017 section 1.

53. Q8. MCQ, answer is C. Gibb’s free energy will be at a minimum
as it has been used up in the reactions and entropy will be at a
maximum as the system can’t be any more disordered.
C

54. Topic 5/15 review
See practice questions☺